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Exercises 4.4 Exercises

1.

The sequence \(2^n\)—that is, the powers of \(2\)—can be represented recursively in two different ways. What is one of those ways?

Hint.

Hint: One of them involves addition, the other involves multiplication.

2.

Write \(2391\) in binary.

3.

Write \((11110101010)_2\) in decimal.

4.

Write \(954\) in binary.

5.

Write \(213\) in octal.

6.

Write \(756\) in hexadecimal.

7.

Write \((AF3)_{16}\) in decimal.

8.

Use binary numbers to prove that

\begin{equation*} 1 + 2 + 4 + 8 + \cdots + 2^n = 2^{n+1}-1 \end{equation*}
Hint.

Hint: What do you get when you add \(1\) to \((111)_2\text{?}\) What about \((1111)_2\text{?}\)

Solution.

Solution: The number

\begin{equation*} 1+2+4+8+\cdots + 2^n \end{equation*}

can be represented as the \(n+1\) digit number \((111\ldots 11)_2\text{.}\) If we add \(1\) to this number, we will have to “carry the one” several times and the result will be the \(n+2\) digit number

\begin{equation*} (1000\ldots 00)_2 \end{equation*}

which is \(2^{n+1}\text{.}\) So, we have shown

\begin{equation*} 1+2+4+8+\cdots + 2^n + 1 = 2^{n+1} \end{equation*}

which can be rearranged to

\begin{equation*} 1+2+4+8+\cdots + 2^n = 2^{n+1}-1 \end{equation*}
9.

A number written in ternary is written in base-\(3\) with digits \(0\text{,}\) \(1\text{,}\) and \(2\text{.}\) Write the number \(143\) in ternary.

10.

Write the number \(143\) in base-\(9\text{.}\)